php-db | 2001071

Re: [PHP-DB] a simple question From: Frank M. Kromann (frank <email protected>)
Date: 07/09/01

Hi Brad,

Your parameter is named $title and you are converting a variable called $input to an array.

You could catch this type of errors if you change the setting for error_reportiong in php.ini.

- Frank

> can somebody help me with the following code:
>
>
> function multi_word($title) {
>
> $array = explode (" ", $input);
> $number = count ($array);
>
> if ($number > 1) {
> return true;
> } else{
> return false;
> }
>
> }
>
> if ($category) {
> $Query = "SELECT * from $TableName WHERE ((category)=$category)";
> } else {
> if ($title) {
> if (multi_word($title)) {
> print "one word at a time for now";
> die();
> } else {
> $Query = "SELECT * from $TableName WHERE title LIKE '%$title%'";
> }
> }
> }
>
> when i try to pass in a multi-word string it doesnt print anything. I have
> tested the function and I know it works allright... if anyone could help
> that would be great.
>
> Brad
>
>
>
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